# Complex function with permutation

Consider the function $$f:C^2\rightarrow C$$ and it is defined by:

$$f(z_a,z_b)=R(z_a)+iI(z_b)$$

where $$R(z_a)$$ is the real part of $$z_a$$ and $$I(z_b)$$ is the imaginary part of $$z_b$$.

Can I write $$f(z_a,f(z_b,z_c))=f(f(z_a,z_b),z_c)$$ in this case?

I spotted where I went wrong.

My working:

$$f(z_a,f(z_b,z_c))=R(z_a)+iI(z_c)=f(f(z_a,z_b),z_c)$$

I have accidentally put down $$f(z_a,f(z_b,z_c))=R(z_a)+iI(z_b)$$ but still got $$f(f(z_a,z_b),z_c)=R(z_a)+iI(z_c)$$ in my previous working so my LHS is not same as RHS and could not find out the mistake by reading thought my handwriting. But once I typed it up, I spotted it.

So this equation is correct.

• That property is called "associativity". Also, here on MSE it's very good to show your work and where you got stuck. – Richard Nov 1 '18 at 18:04
• Thanks for your advice, and I literually spotted my mistake by typing my working up. Thank! – BlackSky Nov 1 '18 at 18:16